HDU 2844 Coins

像下面代码直接利用二进制求解多重背包会超时

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <vector>
#include <algorithm>

using namespace std;

int main(){
    int n,m;
    while(cin>>n>>m && n && m){
        vector<int> A(n),C(n);
        for(int i = 0 ; i < n; i ++ ) cin >> A[i];
        for(int i = 0 ; i < n; i ++ ) cin >> C[i];
        vector<int> coin;

        for(int i = 0; i < n; i ++ ){
            int tmp = 1;
            while(C[i] > tmp){
                coin.push_back(tmp*A[i]);
                C[i] -= tmp;
                tmp <<=1;
            }
            coin.push_back(C[i]*A[i]);
        }

        vector<int> dp(m+1,0);
        for(int i = 0 ; i < coin.size(); i ++ ){
            for(int j = m ; j >= coin[i]; j -- ){
                dp[j] = max(dp[j],dp[j-coin[i]] + coin[i]);
            }
        }

        int cnt = 0;
        for(int i = 1; i <= m ; i++)
            if(dp[i] != dp[i-1]) cnt++;
        cout<<cnt<<endl;
    }
    return 0;
}

参考背包九讲，对多重背包进行优化

#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>

#define MAX 100000
using namespace std;

int dp[MAX];

void CompletePack(int val, int m){
    for(int j = val; j <= m; j ++ )
        dp[j] = max(dp[j],dp[j - val]+ val);
}

void ZeroOnePack(int val, int m){
    for(int j = m; j >= val; j -- )
        dp[j] = max(dp[j],dp[j - val]+ val);
}

void MultiplePack(int val,int num, int m){
    if(val*num >= m ){          //对于固定的m来说，相当于有无限个硬币，
        CompletePack(val,m); //完全背包
        return;
    }
    int k = 1;               //利用二进制优化转化为01背包
    while( k < num){
        ZeroOnePack(k*val,m);
        num -= k;
        k <<= 1;
    }
    ZeroOnePack(num*val,m);
}

int main(){
    int n,m;
    while(cin>>n>>m && n && m){
        vector<int> A(n),C(n);
        for(int i = 0 ; i < n; i ++ ) cin >> A[i];
        for(int i = 0 ; i < n; i ++ ) cin >> C[i];
        //vector<int> dp(m+1,0);
        memset(dp,0,sizeof(dp));
        for(int i=0; i < n; i ++ )
            MultiplePack(A[i],C[i],m);
        int cnt = 0;
        for(int i = 1; i <= m ; i++)
            if(dp[i] == i) cnt++;
        cout<<cnt<<endl;
    }
    return 0;
}